Calculate $\Delta G$ and $E_{cell}$ for the following cell at $298 \ K$ temperature.
$Al_{(s)} | Al^{3+} (0.01 \ M) || Fe^{2+} (0.02 \ M) | Fe_{(s)}$ $\left[ E^o_{Al^{3+}|Al} = -1.66 \ V \right.$ and $\left. E^o_{Fe^{2+}|Fe} = -0.44 \ V \right]$

(N/A) Calculation of $E^o_{cell}$ :
$E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o_{Fe^{2+}|Fe} - E^o_{Al^{3+}|Al}$
$= [-0.44 - (-1.66)] \ V = 1.22 \ V$
Calculation of $E_{cell}$ :
For the cell reaction: $2Al_{(s)} + 3Fe^{2+}(0.02 \ M) \rightarrow 2Al^{3+}(0.01 \ M) + 3Fe_{(s)}$,$n = 6$.
Using Nernst equation: $E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}$
$E_{cell} = 1.22 - \frac{0.0591}{6} \log \frac{(0.01)^2}{(0.02)^3} = 1.22 - 0.00985 \log \left( \frac{10^{-4}}{8 \times 10^{-6}} \right)$
$E_{cell} = 1.22 - 0.00985 \log (12.5) = 1.22 - 0.00985 \times 1.0969 \approx 1.209 \ V$
Calculation of $\Delta G$ :
$\Delta G = -nFE_{cell} = -6 \times 96500 \times 1.209 = -700017 \ J \ mol^{-1} \approx -700 \ kJ \ mol^{-1}$